AcWing 796. 子矩阵的和
这个题的重点是仿照一维的数组，所以a[N][N]也是从1索引开始的。画个图举个例子就非常清晰了
之所以不好理解是因为没画格子，一个格子代表一个点，就很好理解了。
java代码：
import java.io.*;
public class Main{static int N = 1010;static int[][] arr = new int[N][N];static int[][] s = new int[N][N];public static void main(String[] args) throws IOException{BufferedReader br = new BufferedReader(new InputStreamReader(System.in));String[] str1 = br.readLine().split(" ");int n = Integer.parseInt(str1[0]);int m = Integer.parseInt(str1[1]);int q = Integer.parseInt(str1[2]);for(int i = 1; i <= n; i++){String[] str2 = br.readLine().split(" ");for(int j = 1; j <= m; j++){arr[i][j] = Integer.parseInt(str2[j-1]);}}for(int i = 1; i <= n; i++){for(int j = 1; j <= m;j++){s[i][j] = arr[i][j] + s[i-1][j] + s[i][j-1] - s[i-1][j-1];}}for(int i = 0 ; i < q ; i++){String[] str3 = br.readLine().split(" ");int x1 = Integer.parseInt(str3[0]);int y1 = Integer.parseInt(str3[1]);int x2 = Integer.parseInt(str3[2]);int y2 = Integer.parseInt(str3[3]);System.out.println(s[x2][y2] - s[x1 - 1][y2] - s[x2][y1 - 1] + s[x1 - 1][y1 - 1]);}}
}